Let $f$ be a function defined for all real numbers except for $0$. Also let $f'$, the derivative of $f$, be defined as $f'(x)=\dfrac{(x-2)^3}{x}$. On which intervals is $f$ increasing? Choose 1 answer: Choose 1 answer: (Choice A) A $(-\infty,0)$ and $(2,\infty)$ (Choice B) B $(-\infty,0)$ only (Choice C) C $(-\infty,0)$ and $(0,2)$ (Choice D) D $(0,2)$ and $(2,\infty)$ (Choice E) E The entire domain of $f$
Explanation: We can analyze the intervals where $f$ is increasing/decreasing by looking for the intervals where its derivative $f'$ is positive/negative. A function can only change its direction from increasing to decreasing and vice versa between its critical points and the points where the function itself is undefined. We are given that $f'(x)=\dfrac{(x-2)^3}{x}$ and that $f$ is undefined at $x=0$. $f'(x)=0$ for $x=2$. Our critical point is $x=2$, and we should also consider $x=0$. Our points divide the number line into three intervals: $\llap{-}1$ $0$ $1$ $2$ $3$ $(-\infty,0)$ $(0,2)$ $(2,\infty)$ Let's evaluate $f'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $f'(x)$ Verdict $(-\infty,0)$ $x=-1$ $f'(-1)=27>0$ $f$ is increasing $\nearrow$ $(0,2)$ $x=1$ $f'(1)=-1<0$ $f$ is decreasing $\searrow$ $(2,\infty)$ $x=3$ $f'(3)=\dfrac13>0$ $f$ is increasing $\nearrow$ In conclusion, $f$ is increasing over the intervals $(-\infty,0)$ and $(2,\infty)$.